An observer finds that the elevation of the top of a tower is `22.5^@` and after walking100 m towards the foot of the tower, he finds that the elevation of the top has increased to `67.5^@.` The height of the tower in metres is:

Options

A. 50
B. 75
C. 125
D. 175

Answer

To solve the problem, we will use trigonometric principles, specifically the tangent function, which relates the angles of elevation to the height of the tower and the distances involved.

### Step-by-Step Solution

1. **Understanding the Problem:**
- Let the height of the tower be \( h \) meters.
- The observer is initially at point \( C \) and sees the top of the tower at an angle of elevation of \( 22.5^\circ \).
- After walking 100 meters towards the tower, he reaches point \( D \) and sees the top at an angle of elevation of \( 67.5^\circ \).

2. **Setting Up the Diagram:**
- Let \( AB \) be the tower, \( C \) the initial position of the observer, and \( D \) the position after walking 100 meters.
- Let \( BC \) be the distance from the observer to the foot of the tower initially, and \( BD \) be the distance from the foot of the tower to point \( D \) (which is 100 m).

3. **Using the Tangent Function:**
- From triangle \( ABC \) (the larger triangle), we have:
\[ \tan(22.5^\circ) = \frac{h}{BC} \]
Therefore,
\[ h = BC \cdot \tan(22.5^\circ) \quad \text{(Equation 1)} \]

- From triangle \( ABD \) (the smaller triangle), we have:
\[ \tan(67.5^\circ) = \frac{h}{BC - 100} \]
Therefore,
\[ h = (BC - 100) \cdot \tan(67.5^\circ) \quad \text{(Equation 2)} \]

4. **Substituting Values of Tangent:**
- We know:
\[ \tan(22.5^\circ) = \sqrt{2} - 1 \]
\[ \tan(67.5^\circ) = \sqrt{2} + 1 \]
- Substituting these values into the equations:
- From Equation 1:
\[ h = BC \cdot (\sqrt{2} - 1) \quad \text{(Equation 1)} \]
- From Equation 2:
\[ h = (BC - 100) \cdot (\sqrt{2} + 1) \quad \text{(Equation 2)} \]

5. **Equating the Two Expressions for \( h \):**
\[ BC \cdot (\sqrt{2} - 1) = (BC - 100) \cdot (\sqrt{2} + 1) \]

6. **Expanding and Rearranging:**
\[ BC(\sqrt{2} - 1) = BC(\sqrt{2} + 1) - 100(\sqrt{2} + 1) \]
\[ BC(\sqrt{2} - 1 - \sqrt{2} - 1) = -100(\sqrt{2} + 1) \]
\[ BC(-2) = -100(\sqrt{2} + 1) \]
\[ BC = \frac{100(\sqrt{2} + 1)}{2} = 50(\sqrt{2} + 1) \]

7. **Finding the Height \( h \):**
- Substitute \( BC \) back into Equation 1:
\[ h = 50(\sqrt{2} + 1)(\sqrt{2} - 1) \]
- Using the difference of squares:
\[ h = 50((\sqrt{2})^2 - 1^2) = 50(2 - 1) = 50 \text{ meters} \]

### Final Answer:
The height of the tower is \( 50 \) meters.